Let's take this definition of a normal subgroup:
A subgroup, $N$, of a group, $G$, is called a normal subgroup if it is invariant under conjugation; that is, the conjugation of an element of $N$ by an element of $G$ is still in $N$: $\displaystyle N\triangleleft G\,\Leftrightarrow \;\forall \;n\in N,\;\forall \ g\in G\colon \;gng^{-1}\in N.$
Intuitively one could think that $gng^{-1} \Leftrightarrow g^{-1}gn = n\cdot1 $
But I assume, that this is not true for all $g \in G$.
Can you please provide a simple example for such a $g \in G$, so that $gng^{-1}\not = n$ ?
In fact $gng^{-1}=n$ iff $gn=ng$ i.e. $g$ and $n$ commute. Then you can take $2\times 2$ matrices or permutations on 3 objects.
$$ g= \begin{pmatrix} 1& 1\\ 0 & 1 \end{pmatrix}\ ;\ n= \begin{pmatrix} 1& 0\\ 1 & 1 \end{pmatrix} $$ hence $$ gng^{-1}= \begin{pmatrix} 1& 1\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1& 0\\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1& -1\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2& -1\\ 1 & 0 \end{pmatrix} \not=n $$ For permutations, take $g=(1,2)=g^{-1}$ (exchange of $1$ and $2$) and $n=(2,3)$ (exchange of $2$ and $3$). Composing the permutations (as functions), one has $$ gng^{-1}(2)=gn(1)=g(1)=2\ ;\ n(1)=1 $$
thus $gng^{-1}\not=n$. Hope it helps !